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Time resolution of Spectral Entropy - How could I modify it?

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John Navarro
John Navarro on 10 Mar 2021
Edited: John Navarro on 10 Mar 2021
Hello everyone,
I have a signal (a timetable of 307,200 data points) with sampling rate of 20480 Hz (0.00005s) and total length of 15.0 seconds.
When I apply the command pentropy I get a time vector te with length 500 points, equivalent to a time resolution of 0.03 seconds
I confirmed this value as shown below. The problem is that I need the spectral entropy of the signal every 0.02 seconds and every 0.05 seconds.
Is any way where I can adjust or define this "time resolution"?
[se,te] = pentropy(Datos01.Sensor1,Fs)
mean(diff(te))
ans =
0.02997
Could someone help me? Thanks
P.S. I was planning to use the retime command and @pentropy as input but it sends me an error. See Below
Other alternative would be using a for and a moving window but not sure how to code it
DatosA = retime(Datos01,"regular",@pentropy,"TimeStep",seconds(windowLength));
% Error using timetable/retime (line 140)
% Applying the function 'pentropy' to the 1st group in the variable 'Sensor01' generated the following error:
% Expected input argument 2 to be time information in the form of a numeric scalar as sampling frequency, a duration
% scalar as sampling time or a numeric/duration/datetime array as time values.

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